quadratic equations Model Questions & Answers, Practice Test for ibps so prelims 2023

Answer: (d)

Given equation $px^2$ + qx + r = 0.

Let one root of the equation be α.

So, other roots = 3 α

∴ Sum of roots = α + 3 α = - $q/p$

⇒ 4 α = ${-q}/p ⇒ α = {-q}/{4p}$.......(i)

Product of roots = (α).(3 α) = $r/p$

Put the value of α

⇒ $3 α^2 = r/p ⇒ 3({-q}/{4p})^2 = r/p$

⇒ ${3q^2}/{16p^2} = r/p ⇒ 3q^2 p = 16p^2 r$

⇒ $3q^2$ = 16pr

Answer: (b)

f(x) = $2x^3 + x^2$ - 2x - 1

= $x^2(2x + 1) - 1(2x + 1) = (2x + 1) (x^2 - 1)$

= (2x + 1) (x + 1) (x - 1)

So given expression is divisible by 2x + 1.

Answer: (c)

$x^2 = 6 + √{6 + √{6 + √{6 + .....∞}}}$

$x^2 = 6 + √{x^2}$

⇒ $x^2$ = 6 + x

⇒ $x^2$ - x - 6 = 0

⇒ $x^2$ + 2x - 3x - 6 = 0

⇒ x(x + 2) - 3(x + 2) = 0

⇒ (x - 3) (x + 2) = 0

∴ x = 3

Alternate Method:

Given,

$x^2 = 6 + √{6 + √{6 + √{6 + ..... ∞}}}$

x = $√{6 + √{6 + √{6 + √{6 + ......∞}}}}$

Here, factor of 6 = 2 and 3

Sign in expression is positive.

Sothat x = 3.

Answer: (e)

Note: Let the quardatic equation be $ax^2$ + b + c = 0.

To find roots of this equation quickly, we find two factors of 'b' such that their sum is equal to b and their product is equal to the product of the coefficient of x2 and the constant term 'c'.

Let two such factors be α and β.

The α + β = b and α β = ca

In the second step, we divide these factors by the coefficient of $x^2$ ,

ie be 'a'.

In the next step, we change the signs of the outcome. These are the

roots of the equation.

Answer: (e)

I. ${3}/{√x} + {4}/{√x} = √x⇒{7}/{√x} = √x$

⇒7 = $√{x^2}$ = x ∴ x = 7

II. $y^2 - {(7)^{5/2}}/{√y} = 0⇒ y^2 = {(7)^{5/2}}/{√y}$

⇒$7^2 × √y = (7)^{5/2}⇒(y)^{5/2} = (7)^{5/2}$⇒y = 7